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# 17.15: Hess's Law of Heat Summation

Difficulty Level: At Grade Created by: CK-12
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Practice Hess's Law of Heat Summation
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#### How much energy is involved in the operation of an acetylene torch?

Since there is a complex series of reactions taking place, simple methods for determining the heat of reaction will not work. We need to develop new approaches to these calculations.

It is sometimes very difficult or even impossible to measure the enthalpy change for a reaction directly in the laboratory. Some reactions take place extremely slowly, making a direct measurement unfeasible. In other cases, a given reaction may be an intermediate step in a series of reactions. Some reactions may be difficult to isolate because multiple side reactions may occur at the same time. Fortunately, it is possible to measure the enthalpy change for a reaction by an indirect method. Hess’s law of heat summation states that if two or more thermochemical equations can be added together to give a final equation, then the heats of reaction can also be added to give a heat of reaction for the final equation.

An example will illustrate how Hess’s law can be used. Acetylene (C2H2) is a gas that burns at an extremely high temperature (3300°C) and is used in welding (pictured in the opening image). On paper, acetylene gas can be produced by the reaction of solid carbon (graphite) with hydrogen gas.

2C(s, graphite)+H2(g)C2H2(g)H=?\begin{align*}2\text{C}(s, \ \text{graphite})+\text{H}_2(g)\rightarrow \text{C}_2\text{H}_2(g) \qquad \triangle \text{H}=?\end{align*}

Unfortunately, this reaction would be virtually impossible to perform in the laboratory because carbon would react with hydrogen to form many different hydrocarbon products simultaneously. There is no way to create conditions under which only acetylene would be produced.

However, enthalpy changes for combustion reactions are relatively easy to measure. The heats of combustion for carbon, hydrogen, and acetylene are shown below along with each balanced equation.

1. C(s, graphite)+O2(g)CO2(g)H=393.5 kJ\begin{align*}\text{C}(s, \ \text{graphite})+\text{O}_2(g)\rightarrow \text{CO}_2(g) \qquad \qquad \quad \triangle \text{H}=-393.5\ \text{kJ}\end{align*}
2. H2(g)+12O2(g)H2O(l)H=285.8 kJ\begin{align*}\text{H}_2(g)+\frac{1}{2}\text{O}_2(g)\rightarrow \text{H}_2\text{O}(l) \qquad \qquad \qquad \qquad \triangle \text{H}=-285.8\ \text{kJ}\end{align*}
3. C2H2(g)+52O2(g)2CO2(g)+H2O(l)H=1301.1 kJ\begin{align*}\text{C}_2\text{H}_2(g)+\frac{5}{2}\text{O}_2(g)\rightarrow 2\text{CO}_2(g) +\text{H}_2\text{O}(l)\qquad \triangle \text{H}=-1301.1\ \text{kJ}\end{align*}

To use Hess’s law, we need to determine how the three equations above can be manipulated so that they can be added together to result in the desired equation (the formation of acetylene from carbon and hydrogen).

In order to do this, we will go through the desired equation, one substance at a time – choosing the combustion reaction from the equations numbered 1-3 above that contains that substance. It may be necessary to either reverse a combustion reaction or multiply it by some factor in order to make it “fit” to the desired equation. The first reactant is carbon and the in the equation for the desired reaction, the coefficient of the carbon is a 2. So, we will write the first combustion reaction, doubling all of the coefficients and the H\begin{align*}\triangle \text{H}\end{align*}.

2C(s, graphite)+2O2(g)2CO2(g)H=2(393.5)=787.0 kJ\begin{align*}2\text{C}(s, \ \text{graphite})+2\text{O}_2(g)\rightarrow 2\text{CO}_2(g) \qquad \triangle \text{H}=2(-393.5)=-787.0\ \text{kJ}\end{align*}

The second reactant is hydrogen and its coefficient is a 1, as it is in the second combustion reaction. Therefore, that reaction will be used as written.

H2(g)+12O2(g)H2O(l)H=285.8 kJ\begin{align*}\text{H}_2(g)+\frac{1}{2}\text{O}_2(g)\rightarrow \text{H}_2\text{O}(l) \qquad \triangle \text{H}=-285.8\ \text{kJ}\end{align*}

The product of the reaction is C2H2 and its coefficient is also a 1. In combustion reaction #3, the acetylene is a reactant. Therefore, we will reverse reaction 3, changing the sign of the H\begin{align*}\triangle \text{H}\end{align*}.

2CO2(g)+H2O(l)C2H2(g)+52O2(g)H=1301.1 kJ\begin{align*}2\text{CO}_2(g)+\text{H}_2\text{O}(l)\rightarrow \text{C}_2\text{H}_2(g)+\frac{5}{2}\text{O}_2(g)\qquad \triangle \text{H}=1301.1\ \text{kJ}\end{align*}

Now, these three reactions can be summed together. Any substance that appears in equal quantities as a reactant in one equation and a product in another equation cancels out algebraically. The values for the enthalpy changes are likewise added.

2C(s, graphite)+2O2(g)H2(g)+12O2(g)2CO2(g)+H2O(l)2C(s, graphite)+H2(g)2CO2(g)H2O(l)C2H2(g)+52O2(g)C2H2(g)H=787.0 kJH=285.8 kJH=1301.1 kJH=228.3 kJ\begin{align*}2\text{C}(s, \ \text{graphite})+\cancel{2\text{O}_2(g)} &\rightarrow \cancel{2\text{CO}_2(g)} &&\triangle \text{H}=-787.0\ \text{kJ}\\ \text{H}_2(g)+\cancel{\frac{1}{2}\text{O}_2(g)} &\rightarrow \cancel{\text{H}_2\text{O}(l)} &&\triangle \text{H}=-285.8\ \text{kJ}\\ \cancel{2\text{CO}_2(g)}+\cancel{\text{H}_2\text{O}(l)} &\rightarrow \text{C}_2\text{H}_2(g) +\cancel{\frac{5}{2}\text{O}_2(g)} &&\triangle \text{H}=1301.1\ \text{kJ}\\ \hline 2\text{C}(s, \ \text{graphite})+\text{H}_2(g) &\rightarrow \text{C}_2\text{H}_2(g)\qquad \qquad && \triangle \text{H}=228.3\ \text{kJ}\end{align*}

So the heat of reaction for the combination of carbon with hydrogen to produce acetylene is 228.3 kJ. When one mole of acetylene is produced, 228.3 kJ of heat are absorbed, making the reaction endothermic.

### Summary

• Hess’ Law is used to calculate the heat of reaction for processes that cannot be measured directly.

### Review

1. List two reasons direct measurement of a heat of reaction may not be feasible.
2. State Hess’ law.
3. If a combustion reaction is reversed, what needs to happen to the H\begin{align*}\triangle \text{H}\end{align*} value?

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